# Extending a uniformly cont function on an open interval to a closed interval?

1. Oct 23, 2011

### moxy

1. The problem statement, all variables and given/known data
Show that the function $f: J → ℝ$ is bounded if $f$ is uniformly continuous on the bounded interval J.

2. Relevant equations
J is a bounded interval, so say J = (a,b)

$f$ is uniformly continuous on J, so
$\forall \epsilon > 0$ there exists a $\delta > 0$ such that for $s,t \in J = (a,b)$
$|f(s) - f(t)| < \epsilon$ whenever $|s - t| < \delta$

$f: J → ℝ$ is bounded if there exists a real number M such that |f(x)| ≤ M for all x in J.

3. The attempt at a solution
I think that if I can extend $f$ to the endpoints of J, then I can use the Extreme Value Theorem to say that $f$ attains a min and max value, i.e. is bounded. So I need to define $f(a)$ and $f(b)$.

$f(a) = \lim_{n→∞}{f(a + \frac{1}{n})}$
$f(b) = \lim_{n→∞}{f(b - \frac{1}{n})}$

Where clearly $f$ is defined on $[a + \frac{1}{n}, b - \frac{1}{n}]$ for all $n \in ℕ$

Am I on the right track? I don't feel like I've used the fact that $f$ is uniformly continuous on J. Is it because $f$ is uniformly continuous that I'm able to define $f(a)$ and $f(b)$?

2. Oct 23, 2011

### deluks917

Try to do this more simply. The idea is to pick x1,....,xn such that every x in J is "close" to one of these xi. This plus uniform continuity will give you what you want.

3. Oct 23, 2011

### moxy

But I only know that J is bounded, not that it's closed and bounded. So can't I only say that J = (a,b) and not necessarily J = [a,b]? So I don't immediately know if f is defined at a or b, much less if it's continuous at a and b.

4. Oct 23, 2011

### moxy

Okay, nevermind, you changed your post. So... I should define a sequence of, say, rational numbers {xn}, where all xi are in J, such that $lim_{n->∞}{\{x_n\}} = a$ ?

And similarly, {yn}, where all yi are in J, such that $lim_{n->∞}{\{y_n\}} = b$ ?

I think that I'm able to do this, but I don't see how it gets me any closer to showing that f is bounded.

EDIT:

I guess I can say that

$lim_{n->∞}{f(x_n)} = f(a)$
$lim_{n->∞}{f(y_n)} = f(b)$

So f is defined on [a,b]. Do I still have to show it's continuous at a and b to use the extreme value theorem? I don't see how this approach is simpler or much different than the one in my original post.

Last edited: Oct 23, 2011